Which diagram shows a light ray moving from a more dense into a less dense medium? A black horizontal line intersected at right
2 answers:
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Answer:
Part a)
Part b)
Explanation:
Two sleds are connected by a rope
mass of each sled is given as
now we know that dog exert pulling force on the rope connected to first sled
Part a)
By newton's first law we know that
Part b)
As we know that force between two ropes will pull the sled behind
so we will have
This question is incomplete, the complete question is;
Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.
Answer:
- the magnitude of the resultant force is 557.32 kN
- the direction of resultant force is 48.29°
Explanation:
Given the data in the question and the diagram below,
First we work on the force on the left hand side.
Left Horizontal
= βgAr
here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²
we substitute
= βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N
Left Vertical
= ( βgπh² / 2 ) × W
we substitute
= [ ( ( 1.6 × 1000 ) × 9.81 × π(1.5)² ) / 2 ] × 6 = 332845.458 N
Now we go to the right hand side
Right Horizontal
= βgAh
here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²
we substitute
= ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N
Right Vertical
= ( βgπh² / 4 ) × W
we substitute
= [ ( ( 0.8 × 1000 ) × 9.81 × π(1.5)² ) / 4 ] × 6 = 83211.36 N
Hence
Fx = -
= 52974 N - 423792 N = 370818 N
Fy = +
= 332845.458 N + 83211.36 N = 416056.818 N
R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N
R = 557.32 kN
Therefore, the magnitude of the resultant force is 557.32 kN
Direction of resultant force;
tanθ = Fy / Fx
we substitute
tanθ = 416056.818 N / 370818 N
tanθ = 1.121997
θ = tan⁻¹( 1.121997 )
θ = 48.29°
Therefore, the direction of resultant force is 48.29°
Answer:
Explanation:
Given data:
v = 220 rms
power factor = 0.65
P = 1250 W
New power factor is 0.9 lag
we knwo that
s = 1923.09 < 49.65^o
s = [1250 + 1461 j] vA
solving for
Faraday